Confusion over distinguished triangle

According to https://stacks.math.columbia.edu/tag/08J5 we have for every complex $$K$$ and integer $$a$$ a distinguished triangle $$\tau_{\leq a}K\rightarrow\tau_{\leq a+1}K\rightarrow H^{a+1}(K)[-a-1]\rightarrow\tau_{\leq a}K.$$ If we consider the associated long exact sequence we get $$0=H^{a+1}(\tau_{\leq a}K)\rightarrow H^{a+1}(\tau_{\leq a+1}K)=H^{a+1}(K)\rightarrow H^{a+1}(H^{a+1}(K)[-a-1])=0$$ and so we should have $$H^{a+1}(K)=0$$, but there is no reason for this to be true. What is going wrong here?

I'm a newb with derived categories, so this may be wrong:

What's the convention used for the shift functors in the stacks project? I think the confusion comes from the fact that we are using cochains in $$K(\mathcal A)$$.

In section 13.9, we are referred back to the following:

https://stacks.math.columbia.edu/tag/011G

So, if $$A^\bullet$$ is a cochain complex in $$\mathcal A$$, the shifted complex is

$$A[k]^n = A^{n+k}$$

So

$$A^\bullet: \cdots \rightarrow A^{-2} \rightarrow A^{-1} \rightarrow A^0 \rightarrow A^1 \rightarrow A^2 \rightarrow \cdots$$

gets shifted to

$$A[k]^\bullet: \cdots \rightarrow A^{-2 + k} \rightarrow A^{-1 + k} \rightarrow A^k \rightarrow A^{1+k} \rightarrow A^{2+k} \rightarrow \cdots$$

That is, $$$$ is a left shift !

That makes sense because the $$C \rightarrow A$$ part of a distinguished triangle should record the map $$C^n \rightarrow A^{n+1}$$ in degree $$n$$.

Now, I believe the convention is that $$H^{a+1}(K^\bullet)$$ denotes the complex

$$\cdots \rightarrow 0 \rightarrow 0 \rightarrow H^{a+1}(K^\bullet) \rightarrow 0 \rightarrow 0 \cdots$$,

with $$(H^{a+1}(K^\bullet))^0 = H^{a+1}(K^\bullet)$$ and $$H^{a+1}(K^\bullet))^n = 0$$ otherwise.

Then, the right shift $$H^{a+1}(K^\bullet)[-a-1]$$ is the complex with

$$(H^{a+1}(K^\bullet)[-a-1])^n = \begin{cases} H^{a+1}(K^\bullet), & n=a+1 \\ 0, & \mbox{otherwise} \end{cases}$$

so then

$$H^n(H^{a+1}(K^\bullet)[-a-1])) = \begin{cases} H^{a+1}(K^\bullet), & n=a+1 \\ 0, & \mbox{otherwise} \end{cases}$$

and the long exact sequence associated to that distinguished triangle is

$$\scriptsize \require{AMScd} \begin{CD} \cdots H^a(\tau_{\leq a}K^\bullet) @>>> H^a(\tau_{\leq a+1} K^\bullet) @>>> H^a(H^{a+1}(K^\bullet)) @>>> H^{a+1}(\tau_{\leq a}K^\bullet) @>>> H^{a+1}(\tau_{\leq a+1} K^\bullet) @>>> H^{a+1}(H^{a+1}(K^\bullet)) @>>> H^{a+2}(\tau_{\leq a} K^\bullet) \cdots \\ @VVV @VV=V @VV=V @VV=V @VV=V @VV=V @VV=V \\ \cdots H^a(K^\bullet) @>>> H^a(K^\bullet) @>>> 0 @>>> 0 @>>> H^{a+1}(K^\bullet) @>>> H^{a+1}(K^\bullet) @>>> 0 \cdots \end{CD}$$

(Sorry if that goes off your screen, I'm not sure how to make arrows shorter in amscd)

• Indeed, you are right. Thanks! For some reason I thought it was a left shift.. – Bernie May 8 at 12:06